Question

The population of a town increased by 10 percent annually. if the present population is 22000,find its population a year ago

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,,

By applying the formula of compound interest we can make the following assumptions and calculate them for the same. You'll come across this formulae.........

Let us take it as "x" years and Total population to find be "Y" and let "PP" be as present population , we'll get this ,,,

$\boxed{\bf{PP = Y \times [1 - \frac{r}{100}]^x}}$

Now, for population changes before "1" years ago, substitute this into "x" years ,,,

$\bf{22000 = Y \times [1 + \frac{10}{100}]^1}$

$\bf{22000 = Y \times [1 + 0.1]^1}$

$\bf{22000 = Y \times [1.1]^1}$

$\bf{22000 = Y \times [1.1]}$

$\boxed{\bf{\therefore \: \: Y = 20000}}$

Therefore the answer is ,,,

$\boxed{Y = 20000}$

HOPE THIS HELPS YOU AND CLEARS THE DOUBTS FOR CALCULATING THE TIMES OF YEAR BY THIS FORMULAE!!!!!!!

Answer 2

HERE IS YOUR ANSWER...⬇⬇

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$\underline{SOLUTION}$ ➡

Given,

Population increased by 10% annually

Present population = 22000

Let ,

The total population of last year = x

A.T.Q.,

x + 10% of x = 22000
$= > x + \frac{10}{100} x = 22000 \\ \\ = > \frac{100x + 10x}{100} = 22000 \\ \\ = > \frac{110x}{100} = 22000 \\ \\ = > 110x = 22000 \times 100 \\ \\ = > 110x = 2200000 \\ \\ = > x = \frac{2200000}{110} \\ \\ = > x = 20000$

•°• Total population of a year ago / last year = 20000

$\underline{ANSWER}$ ➡ 20000

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✝✝...HOPE IT HELPS YOU...✝✝